Hi,
in your example
1 + a + b + c /. a + b -> x
gives
1 + c + x
but for more complicates cases, you have to use
more complicated patterns.
Regards
Jens
Brian Beckage wrote:
> I apologize for this very basic question. I understand how to use /.
> {} to replace a variable with a more complex expression, e.g., x->
> y+z. Can one move in the opposite direction to replace all
> occurrences of y+z with x? myExpression/.{y+z->x} does not seem to
> work.
>
> Thank you for your help,
> Brian
>
>
>